Bodies connected by the cord and pulley – application of Newton’s law of motion problems and solutions

1. Two boxes are connected by a cord running over a pulley. Ignore the mass of the cord and pulley and any friction in the pulley. Mass of the box 1 = 2 kg, mass of the box 2 = 3 kg, acceleration due to gravity = 10 m/s². Find (a) The acceleration of the system (b) The tension in the cord!

Solution

Known :

Mass of the box 1 (m₁) = 2 kg

Mass of the box 2 (m₂) = 3 kg

Acceleration due to gravity (g) = 10 m/s²

Weight of the box 1 (w₁) = m₁ g = (2)(10) = 20 Newton

Weight of the box 2 (w₂) = m₂ g = (3)(10) = 30 Newton

Solution :

(a) magnitude and direction of the acceleration

w₂ > w₁ so the box 2 accelerates downward and the box 1 accelerates upward.

Forces that has the same direction with acceleration (w₂ and T₁), its sign positive. Forces that has opposite direction with acceleration (T₂ and w₁), its sign negative.

∑F = m a

w₂ – T₂ + T₁ – w₁ = (m₁ + m₂) a ——-> T₁ = T₂ = T

w₂ – T + T – w₁ = (m₁ + m₂) a

w₂ – w₁ = (m₁ + m₂) a

30 – 20 = (2 + 3) a

10 = 5 a

a = 10 / 5

a = 2 m/s²

Magnitude of the acceleration is 2 m/s².

(b) The tension force

The box 2 :

There are two forces acts on the box 2 : first, weight of the box 2 (w₂), points downward so it’s positive. Second, tension force exerted on the box 2 (T₂), points upward so it’s negative. Apply Newton’s second law of motion.

∑F = m a

w₂ – T₂ = m₂ a

30 – T₂ = (3)(2)

30 – T₂ = 6

T₂ = 30 – 6

T₂ = 24 Newton

Box 1 :

There are two forces acts on the box 1. First, weight of the box 1 (w₁), points downward so it’s negative. Second, the tension force exerted on the box 1 (T₁) points upward so it’s positive. Apply Newton’s second law of motion :

∑F = m a

T₁ – w₁ = m₁ a

T₁ – 20 = (2)(2)

T₁ – 20 = 4

T₁ = 20 + 4

T₁ = 24 Newton

Magnitude of the tension force = T₁ = T₂ = T = 24 Newton

2. An object on a rough horizontal surface. Mass of the object 1 = 2 kg, mass of the object 2 = 4 kg, acceleration due to gravity = 10 m/s², coefficient of the static friction = 0.4, coefficient of the kinetic friction = 0.3. The system is at rest or accelerated ? If the system is accelerated, find the magnitude and direction of the system’s acceleration!

Solution

Known :

Mass of the object 1 (m₁) = 2 kg

Mass of the object 2 (m₂) = 4 kg

Acceleration due to gravity (g) = 10 m/s²

Coefficient of the static friction (μ_s) = 0.4

The coefficient of the kinetic friction (μ_k) = 0.3

Weight of the object 1 (w₁) = m₁ g = (2)(10) = 20 Newton

Weight of the object 2 (w₂) = m₂ g = (4)(10) = 40 Newton

Normal force exerted on the object 1 (N) = w₁ = 20 Newton

Force of the static friction exerted on the object 1 (f_s) = μ_s N = (0.4)(20) = 8 Newton

Force of the kinetic friction exerted on the object 1 (f_k) = μ_k N = (0.3)(20) = 6 Newton

Wanted: acceleration (a)

Solution :

w₂ > f_s (40 Newton > 8 Newton) so the object 2 is accelerated vertically downward and the object 1 is accelerated horizontally rightward. The friction force that acts on the objects 1 is the force of the kinetic friction (f_k). Apply Newton’s second law of motion :

∑F = m a

w₂ – the = (m₁ + m₂) a

40 – 6 = (2 + 4) a

34 = 6 a

a = 34 / 6 = 17 / 3

a = 5.7 m/s²

Magnitude of the acceleration = 5.7 m/s²

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1. Mass and weight
2. Normal force
3. Newton’s second law of motion
4. Friction force
5. Motion on horizontal surface without friction force
6. The motion of two bodies with the same acceleration on the rough horizontal surface with the friction force
7. Motion on the inclined plane without friction force
8. Motion on the rough inclined plane with the friction force
9. Motion in an elevator
10. The motion of bodies connected by cord and pulley
11. Two bodies with the same magnitude of accelerations
12. Rounding a flat curve – dynamics of circular motion
13. Rounding a banked curve – dynamics of circular motion
14. Uniform motion in a horizontal circle
15. Centripetal force in uniform circular motion